
Optional unit · institution-dependent · Interactive lesson
The postulates and the relativity of simultaneity
Two postulates replace the ether: the same physics in every inertial frame, and one speed of light in all of them. Their first casualty is the phrase 'at the same time'. Learn to synchronise clocks with light, run Einstein's train with real numbers, and compute how far apart two simultaneous events drift in a moving frame.
UPIII-08.02 · Lesson 2 of 9 in this unit. Position is not a completion score.Optional unit · institution-dependent. This modern-physics material is an introductory survey, not a complete course in relativity, quantum, atomic, nuclear or particle physics. Your university determines the required depth and assessment. Optional work is not a required completion gate.
Build the model
Same laws and the same c in every inertial frame, and 'now' stops being shared.
The key idea
Two events are simultaneous in a frame when that frame's light-synchronised clocks, one at each event, read the same time; a frame moving at v along the line joining them assigns the pair a time difference Δt′ = −γvΔx/c², so simultaneity belongs to the frame, not to the events.
Example. A rocket passing at 0.60c (γ = 1.25) over two station beacons 400 m apart that flash together in the station frame finds the flashes 1.25 × 0.60 × 400 m ÷ (3.00 × 10⁸ m s⁻¹) = 1.0 μs apart, the beacon ahead of it flashing first.
The complete physical model and assumptions
Nineteenth-century physics had two relativity principles that did not match. Mechanics obeyed Galileo: no experiment inside a smoothly moving laboratory reveals its speed. Electromagnetism seemed not to, because Maxwell's equations name one speed, c, and a Galilean boost should change it, so Earth's 30 km s⁻¹ through the ether ought to leave a second-order trace of size (v/c)² ≈ 10⁻⁸ in the round-trip time of light. Michelson and Morley built an interferometer sensitive to exactly that and saw nothing. Einstein's response was to take the relativity principle seriously for all of physics and to promote the constancy of c from a property of Maxwell's equations to a law of nature: every inertial observer measures light at c, whatever the motion of source or observer. Everything else in this section follows from those two sentences, starting with a definition. 'Simultaneous' means nothing until you say how clocks at different places are set, and the only honest recipe uses light signals whose speed you already trust. Run that recipe in one frame and in a second moving at v along the line joining two events, and the frames disagree by Δt′ = −γvΔx/c²: not an optical illusion, but what each frame's synchronised clocks record. The postulates govern inertial frames in flat spacetime only. They forbid a preferred frame, fix an invariant speed, and say nothing about gravitation.
1An ether wind should have shifted the fringes
Michelson's interferometer splits a beam into two perpendicular arms of length L and recombines them. If light rode an ether through which Earth moved at v ≈ 30 km s⁻¹, the along-wind arm's round trip would take L/(c − v) + L/(c + v) = (2L/c)γ² and the across-wind arm's 2L/√(c² − v²) = (2L/c)γ, with γ = 1/√(1 − v²/c²). Expanding both, the arms differ by Δt ≈ Lv²/c³. Rotating the apparatus by 90° swaps the roles, so the optical path difference swings by 2cΔt and the pattern walks by N = 2Lv²/(λc²) fringes. For the 1887 instrument, L = 11 m (folded by mirrors), λ = 589 nm and (v/c)² = 1.0 × 10⁻⁸: N = 2 × 11 × 10⁻⁸ ÷ 5.89 × 10⁻⁷ = 0.37 fringe, against a resolution near 0.01. They saw under 0.02, and every more sensitive repetition since has seen nothing. Note the scale of the achievement: the interferometer turns a time difference of 3.7 × 10⁻¹⁶ s into a path difference of 110 nm and compares it with a wavelength, which is how a 10⁻⁸ effect became measurable at all.
2Two postulates, and what each one forbids
Einstein kept the null result and dropped the ether. Postulate 1: the laws of physics take the same form in every inertial frame. Galileo had claimed this for mechanics; extending it to electromagnetism means no experiment, optical or electrical, can detect a laboratory's uniform motion, so there is no ether wind to find. Postulate 2: light in vacuum travels at c in every inertial frame, whatever the motion of its source. Maxwell's equations already contain c = 1/√(ε₀μ₀); the postulate says that number is not tied to a medium but is the same for every inertial observer. Each postulate closes a door. The first removes any preferred frame. The second kills Galilean velocity addition for light: a lamp on a train moving at v does not send light forward at c + v as the platform measures it. Since 1983 the SI has built the second postulate in, defining the metre by fixing c = 299 792 458 m s⁻¹ exactly. Both postulates speak of inertial frames only: they constrain observers in uniform relative motion and say nothing about accelerated frames or gravitation.
3Synchronising clocks with light: what 'the same time' means
'Simultaneous' is not given; it has to be constructed. Two identical clocks at rest in a frame, A at x = 0 and B a distance L away, are synchronised by Einstein's recipe. A sends a flash at its reading t₁; the flash reflects at B and returns to A at t₂. A alone can compute L = c(t₂ − t₁)/2 and the reflection time (t₁ + t₂)/2, and B is set to read that value at the moment the flash arrived. With t₁ = 0 and t₂ = 4.00 μs, B sits 600 m away and is set to 2.00 μs. The recipe uses one clock plus the convention that the outward and return legs take equal times. Repeated for every clock in the frame, it defines the frame's time coordinate, and two events are simultaneous in that frame when the clocks at their locations agree. Postulate 2 makes the recipe consistent: synchronising B from A and then C from B agrees with synchronising C from A directly. Postulate 1 grants every inertial frame the right to run the same recipe with its own clocks. Disagreement is now unavoidable, because each frame trusts light to move at c relative to itself.
4Einstein's train, with the closing speeds done honestly
Platform frame S; a train of S-length L = 300 m moves at v = 0.60c, so γ = 1.25. Lightning strikes both ends at t = 0 in S, meaning S's clocks at x = 0 and x = 300 m both read zero. A passenger sits at the train's midpoint. In S the front flash travels backward while she moves forward, so they close at c + v and meet after L/(2(c + v)) = 300 m ÷ (2 × 1.6c) = 312.5 ns; the rear flash chases her at closing speed c − v and arrives after L/(2(c − v)) = 1250 ns. Closing speeds are legal in S: the light still moves at c, and the passenger moves too. She receives the front flash 937.5 ns of S-time before the rear one. Now switch frames. In S′ she is at rest at the midpoint of her own train, both flashes travelled equal distances at c by postulate 2, and they arrived at different times. The only conclusion open to her is that the front strike happened first. Her clocks at the two ends of the train, synchronised by the light recipe in S′, record exactly that. Neither frame is wrong; each has applied the same definition of simultaneity to its own clocks.
5The offset formula, its sign and its size
The train gives the rule. For two events simultaneous in S and separated by Δx along the direction of S′'s motion, Δt′ = t′₂ − t′₁ = −γvΔx/c², where event 2 lies ahead in the direction S′ travels. The thought experiment supplies the leading-order form −vΔx/c² directly; the exact factor γ enters because the moving frame's rods and clocks are themselves rescaled, which the next two topics derive, and the Lorentz transformation confirms the full formula. The numbers agree: the 937.5 ns arrival gap divided by γ is 750 ns, and γβL/c = 1.25 × 0.60 × 300 m ÷ (3.00 × 10⁸ m s⁻¹) = 750 ns. Check dimensions: (m s⁻¹)(m)/(m² s⁻²) = s. Check the limit: as v/c → 0, γ → 1 and Δt′ → −vΔx/c² → 0, Galileo's shared time. Check the size: a car at 30 m s⁻¹ passing two flashes 100 m apart finds them 3.3 × 10⁻¹⁴ s apart, below any clock. Two rules follow. Events at one place (Δx = 0) that are simultaneous in one frame are simultaneous in all. Separation across the motion never enters; only the component along v matters.
6Where the postulates stop
The postulates constrain inertial frames in flat spacetime, and only those. They do not say accelerated frames are equivalent to inertial ones; an accelerating body is described from an inertial frame, and the passenger who feels the train brake knows she is not inertial, whatever the postulates say about uniform motion. They also say nothing about gravitation. A clock higher in a gravitational potential runs faster than one below, an effect outside special relativity that general relativity supplies; GPS satellite clocks gain about 45 μs per day from altitude and lose about 7 μs per day from speed, and both corrections are applied. Finally, the postulates make c a property of spacetime rather than of light. It is the one speed every inertial observer agrees on, and light travels at it because the photon is massless; had the photon a small mass, the invariant speed would still exist and light would fall short of it. Everything that follows in this section, dilation, contraction, the Lorentz transformation and E = γmc², is derived from these two sentences and inherits their domain.
Key equations
Read the units, assumptions and limits before substituting.
Ether-wind delay between the arms
Δt = (2L/c)(γ² − γ) ≈ L v²/c³
Both round trips are slowed, the along-wind one by the factor γ² and the across-wind one only by γ. The effect is second order in v/c, which is why no first-order optical test had ever seen it.
L arm length (m), v ether-wind speed (m s⁻¹), c (m s⁻¹); γ = 1/√(1 − v²/c²); Δt in sFringe shift on a 90° rotation
N = 2cΔt/λ ≈ 2 L v²/(λ c²)
For L = 11 m, v = 30 km s⁻¹ and λ = 589 nm this predicts 0.37 fringe. Michelson and Morley could resolve 0.01 and saw under 0.02.
λ wavelength (m); N is dimensionless, the number of fringes the pattern walks when the two arms swap rolesEinstein clock synchronisation
tB(at reflection) = t₁ + L/c = ½(t₁ + t₂)
Defines 'the same time' at two places using one clock and light. Every inertial frame runs the identical recipe on its own clocks, which is where the disagreement starts.
t₁ send and t₂ echo-return times on master clock A (s); L distance from A to B (m); B is set to this reading as the flash arrivesFlash-arrival gap at the train's midpoint, in S
tᵣₑₐᵣ − tfront = L v/(c² − v²) = γ² L v/c²
Closing speeds c + v and c − v are legitimate in S: they describe light and passenger approaching each other, not light moving at anything but c.
L train length measured in S (m), v train speed (m s⁻¹); the S times at which the front and rear flashes reach the passengerRelativity of simultaneity
Δt′ = t′₂ − t′₁ = −γ v Δx/c² (for Δt = 0)
The minus sign is the content: the event that lies ahead in S′'s direction of travel happens earlier in S′. As v/c → 0, γ → 1 and Δt′ → 0, Galileo's shared time.
Events simultaneous in S with Δx = x₂ − x₁ along the motion (m); S′ moves at +v (m s⁻¹); γ = 1/√(1 − v²/c²); Δt′ in sMoving clocks out of step
t′ᵣₑₐᵣ − t′front = v L₀/c²
The same fact seen from the platform: the trailing clock leads. Nothing is wrong with the clocks; S is comparing them at what S, not S′, calls one instant.
L₀ proper separation (m) of two clocks synchronised in their own frame S′; the readings are compared at one instant of SInteractive lab
Predict first. Explore the model. Compare the result.
Explore the model
Try this. Leave the delay at 0 and raise β: the front flash reaches the passenger earlier and earlier, and the dashed S′ line tilts so the front strike falls below it, earlier in S′. Then set β back to 0.50 and raise the delay until the two arrival dots merge: S calls the strikes γ²βL/c apart just when S′ calls them simultaneous.
Live results
- LORENTZ FACTOR γ
- 1.155
- FRONT FLASH LEAD AT PASSENGER
- 222.4 ns
- S′ GAP, FRONT − REAR
- -192.6 ns
- S DELAY FOR S′ SIMULTANEITY
- 222.4 ns
Results follow this lesson’s model and assumptions. Displayed values are rounded.
Predict → change one variable → check
Plan a change before running it. Every other input stays at the saved baseline.
Saved baseline. Train speed β = v/c: 0.50; Train length L in S: 100 m; Front-strike delay in S: 0 ns
This experiment is kept only while this page is open. It does not change saved course progress.
Catch the common trap
'The passenger sees the front flash first because she is moving toward it, so once she allows for the travel time the strikes are simultaneous for her too.' That is the Galilean instinct, and postulate 2 forbids it. Allowing for travel time is exactly what she does: she sits at the midpoint of her own train, both flashes covered the same distance, and both moved at c relative to her, so unequal arrival means unequal departure. 'Moving toward it' is a platform-frame description that gives the light a closing speed c + v relative to her; in her frame the light approaches at c and she is not moving at all. Her conclusion is not a perception but what her light-synchronised clocks at the two ends record. Nor does the platform hold the truth: its simultaneity is the verdict of its own clocks, set by the same recipe.
Practice & worked examples
3 authored examples. Make an attempt, then reveal the reasoning one step at a time.
EasyExample 1
A tabletop Michelson interferometer has arms of L = 1.20 m and uses a helium-neon laser, λ = 633 nm. Taking Earth's orbital speed v = 3.0 × 10⁴ m s⁻¹ as the ether wind, find the time lag between the arms that the ether theory predicts, the fringe shift on rotating the apparatus by 90°, and the arm length that would make the shift one full fringe.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 4 steps revealed.
- (v/c)² = (3.0 × 10⁴ ÷ 3.00 × 10⁸)² = 1.0 × 10⁻⁸. The effect is second order, so the small-β form of the delay is exact to one part in 10⁸.
- Arm lag: Δt ≈ Lv²/c³ = L(v/c)²/c = 1.20 × 1.0 × 10⁻⁸ ÷ 3.00 × 10⁸ = 4.0 × 10⁻¹⁷ s, an optical path difference cΔt = 12 nm.
- Rotating by 90° swaps the arms, so the path difference swings by 2cΔt = 24 nm: N = 24 nm ÷ 633 nm = 0.038 fringe, or directly N = 2L(v/c)²/λ = 2 × 1.20 × 10⁻⁸ ÷ 6.33 × 10⁻⁷ = 0.038.
- One full fringe needs 2L(v/c)²/λ = 1, so L = λ/(2(v/c)²) = 6.33 × 10⁻⁷ ÷ (2 × 10⁻⁸) = 31.7 m. Michelson and Morley folded their beam to 11 m and still expected 0.37 fringe.
Answer. Δt ≈ 4.0 × 10⁻¹⁷ s (12 nm of path); N ≈ 0.038 fringe on rotation; one full fringe would need L ≈ 32 m of arm.
MediumExample 2
In a laboratory, clock A sits at x = 0. (a) A pings an unknown reflector: the flash leaves when A reads 2.00 μs and the echo returns at 12.00 μs. Assign the reflection a position and a lab-frame time. (b) A second reflector at x = 600 m is pinged at 5.00 μs and its echo returns at 9.00 μs. Are the two reflections simultaneous in the lab frame? (c) An observer at A sees the two reflection flashes arrive 3.00 μs apart. Why is that not a contradiction?
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 4 steps revealed.
- (a) Out and back at c take equal times by Einstein's convention, so the reflector is at x = c(t₂ − t₁)/2 = 3.00 × 10⁸ × 10.00 × 10⁻⁶ ÷ 2 = 1500 m.
- The reflection happened halfway through the round trip: t = (t₁ + t₂)/2 = (2.00 + 12.00)/2 = 7.00 μs. A clock at 1500 m synchronised with A would read 7.00 μs at that moment.
- (b) Check the position first: c(9.00 − 5.00) × 10⁻⁶ ÷ 2 = 600 m, consistent with where it was placed. Reflection time (5.00 + 9.00)/2 = 7.00 μs. Both reflections carry lab time 7.00 μs, so they are simultaneous in the lab frame.
- (c) Light from the 1500 m reflection reaches A at 7.00 + 1500/c = 7.00 + 5.00 = 12.00 μs; light from the 600 m reflection at 7.00 + 2.00 = 9.00 μs. The 3.00 μs gap is the extra 900 m of light travel, which is exactly what the synchronisation recipe subtracts. Seeing later is not happening later.
Answer. (a) x = 1500 m at t = 7.00 μs; (b) yes, both at t = 7.00 μs; (c) the 3.00 μs difference is 900 m of light travel, which the frame's time coordinate already removes.
HardExample 3
A probe of proper length 240 m passes a space station at 0.80c. Clocks at its nose and tail were synchronised in the probe frame. (a) In the station frame, by how much and in which sense are the two probe clocks out of step? (b) Two station beacons 150 m apart along the probe's track flash simultaneously in the station frame. Find the time between the flashes in the probe frame and which flashed first. (c) Repeat (b) for a cart passing at 30 m s⁻¹.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- γ = 1/√(1 − 0.80²) = 1/√0.36 = 1/0.60 = 1.67; keep 5/3 in the working.
- (a) Clocks synchronised in their own frame and separated by proper length L₀ are read, from a frame in which they move, as out of step by vL₀/c² = 0.80 × 240 m ÷ (3.00 × 10⁸ m s⁻¹) = 0.64 μs, the tail clock ahead of the nose clock. Cross-check: in the station frame the probe is 240/γ = 144 m long, and γvΔx/c² with Δx = 144 m gives the same 0.64 μs.
- (b) The beacons are simultaneous in the station frame S with Δx = 150 m, and the probe frame S′ moves at 0.80c: Δt′ = −γvΔx/c² = −(5/3) × 0.80 × 150 m ÷ (3.00 × 10⁸ m s⁻¹) = −200 m ÷ (3.00 × 10⁸ m s⁻¹) = −0.667 μs.
- The minus sign puts the beacon ahead of the probe, in its direction of travel, first: the probe records that beacon flashing 0.667 μs before the one behind it. The leading-order estimate −vΔx/c² = −0.400 μs is 40% short; at β = 0.80 the factor γ is not optional.
- (c) For v = 30 m s⁻¹, γ − 1 ≈ 5 × 10⁻¹⁵, so Δt′ ≈ −vΔx/c² = −30 × 150 ÷ (9.00 × 10¹⁶) = −5.0 × 10⁻¹⁴ s. The cart agrees the flashes were simultaneous to fifty femtoseconds: the Galilean world, recovered as the limit.
Answer. (a) 0.64 μs, tail clock ahead of the nose clock; (b) 0.667 μs, the beacon ahead of the probe flashing first; (c) 5.0 × 10⁻¹⁴ s, simultaneous for every practical purpose.
Quick check
Select an answer and explain it before checking.
Review & continue
Two events are simultaneous in a frame when that frame's light-synchronised clocks, one at each event, read the same time; a frame moving at v along the line joining them assigns the pair a time difference Δt′ = −γvΔx/c², so simultaneity belongs to the frame, not to the events.
- Explain the relationship without looking at the equation.
- Check a sign, a unit and a model assumption in your working.
- Return to the lab or worked example if a step is still unclear.
Optional supporting lessons
These chapter resources can help with earlier ideas; they are not additional required Physics III completion tasks.